Z = + 0.28 (0.611 is approximately the proportion in the larger area of the curve associated with a Z score of +0.28)

Insert a standard normal probability table with values of Z from 0.0 to 0.40

Probit Transformation

22 out of 36 juveniles succeeded

with 8 months of supervision (61.1%)

Standard Normal Curve

Area = 0.611

Z = 0.00

Z = + 0.28

Probit & Logit Case Studies of Probation Outcome Study

Model 1 Probit response model

Success = f (months of supervision)

Model 2 Log transform Probit model

Success = f (log months)

Model 3 Logit response model

Success = f (months of supervision)

Model 4 Log transform Logit model

Success = f (log months)

Model 5 Probit response model

Success = f (months & age)

Model 6 2 Probit response models

Male Success = f (months)

Female Success = f (months)

Model 7 2 Probit multivariate models

Male Success = f (months & age)

Female Success = f (months & age)

Model 1

The Effect of the Number of Months of Supervision on Probation Success

Probit = a + b (months)

Probit = -0.70946 + 0.11582 (months)

Probit Model of the Effect of Months of Supervision on Probation Success

Months of Supervision

Total Number of Juveniles

Number Successful

6

38

18 (0.47)

7

42

23 (0.55)

8

36

22 (0.61)

9

30

19 (0.63)

10

48

32 (0.67)

11

45

32 (0.71)

12

40

30 (0.75)

Values in parentheses are the proportions of juveniles who were successful.

Conversion of the Proportion Successful to Associated Z Scores

Months of Supervision

Total Number of Juveniles

Number Successful (p)

Probit Z Score

6

38

18 (0.47)

-0.08

7

42

23 (0.55)

+0.13

8

36

22 (0.61)

+0.28

9

30

19 (0.63)

+0.33

10

48

32 (0.67)

+0.44

11

45

32 (0.71)

+0.55

12

40

30 (0.75)

+0.67

Plot of the Number of Successful Probationers as a Function of

Months of Supervision

Compare this scatterplot with the Probit transform plot on the following exhibit.

Plot of Probit Z Scores as a Function of Months of Supervision

Notice that the conversion of the proportion of successful juvenile
s to Probit Z scores results in a linear relationship over time.

As months of supervision increases, the value of the Probit increases, i.e. the greater the likelihood of being successful on probation.

In Probit analysis, the equation of the best fit line is achieved by maximum likelihood estimation.

What is the Linear Relationship Between Months of Supervision and Probation Success?

The Probit model

Probit = a + b (months)

Estimated best fit linear equation

Probit = -0.70946 + 0.11582 (months)

Interpretation

For every one-month increase in the amount of supervision, the Probit of success (Z score) increases by 0.11582.

Results of the Probit Analysis

Parameter estimates converged after 9 iterations.

Optimal solution found.

Parameter Estimates (PROBIT model: (PROBIT(p)) = Intercept + BX):

Regression Coeff. Standard Error Coeff./S.E.

MONTHS .11582 .03862 2.99914

Intercept Standard Error Intercept/S.E.

-.70946 .35511 -1.99788

Pearson Goodness-of-Fit Chi Square = .180 DF = 5 P = .999

Since Goodness-of-Fit Chi square is NOT significant, no heterogeneity

factor is used in the calculation of confidence limits.

Probit Model

Probit = a + b (month)

a = intercept = -0.70946 (t = -1.99788, df = N – k = 5, p< 0.05)

b = Probit regression coefficient = 0.11582

(t = 2.99914, df = N – k = 5, p< 0.05)

N = 7 months, k = 2 = number of parameters estimated (a & b)

Probit Model Predictions of the Effect of Months of Supervision on Success

Probit = -0.70946 + 0.11582 (months)

Example: Predicted success of juveniles with 8 months supervision

Total = 36, 22 succeeded or 61.1%

Probit = -0.70946 + 0.11582 ( 8 months)

Probit = +0.2171

Z of +0.2171 has an associated area under

the larger portion of the curve of 0.5859

(36 juveniles) (0.5859) = 21.09 successes

Residual = (22 – 21.09) = 0.91

Probit Transformation

22 out of 36 juveniles succeeded with 8 months of supervision (61.1%). The Probit equation yields a Z = +0.2171. Since Z is positive, the area in the larger portion of the curve is 0.5859, or a prediction of a 58.59 % success rate.

Standard Normal Curve

Area = 0.5859

Z = 0.00

Z = + 0.2171

Scatterplot of the Residuals Produced By the Probit Model

Probit Residuals

0.00

1 2 3 4 5 6 7 8 9 10

Months of Supervision

Chi Square Goodness of Fit Test

l2 = S { (residual)2 / [ ni Pi (1-Pi) ] }

Residual = (observed – expected successes)

nI = Number of cases in the group, i.e. level of IV

PI = Predicted proportion of cases (cf. the next exhibit for a summary of the predictions and the residuals)

Sometimes it is of interest to compare the Probit equations for multiple groups to determine if there are significant differences in the Probit coefficients (i.e. slopes) predictor variables across groups.

Examples of different groups

Racial/ethnic groups

Offense groups

Gender groups

Q Are there significant differences between boys and girls in the effect of the number of months of supervision on probation success?

The Concept of Testing the Parallelism of the Slopes of Multiple Group Probit Equations

Step 1

Estimate separate Probit equations for boys and girls on the effect of months of supervision on probation success.

Step 2

Conduct a chi-square test to determine whether the slopes (coefficients) of the two equations differ significantly, i.e.

Whether the probit coefficients of the two equations differ significantly.

Step 3

If the slopes do not differ significantly, compute a common slope to be used for each group

Plot of the Probits of Probation Success for Boys and Girls
Q Is the linear relationship between of months of supervision on success the same (parallel) for male and female juvenile probationers?

Results of the Probit Analysis for

Boys and Girls

Parameter estimates converged after 8 iterations.

Optimal solution found.

Parameter Estimates (PROBIT model: (PROBIT(p)) = Intercept + BX):

Regression Coeff. Standard Error Coeff./S.E.

MONTH .11726 .03922 2.98946

Intercept Standard Error Intercept/S.E. GENDER

-1.01139 .37197 -2.71905 1

-.41904 .36667 -1.14281 2

Pearson Goodness-of-Fit Chi Square = 42.756 DF = 39 P = .313

Parallelism Test Chi Square = .000 DF = 1 P = 1.000

Since Goodness-of-Fit Chi square is NOT significant, no heterogeneity

factor is used in the calculation of confidence limits.

The parallelism test found no significant difference between the slopes of the two equations, so a common slope was computed to be used in each equation, i.e. bC = 0.11726.

Probit male = -1.001139 + 0.11726 (month)

Probit female = -0.41904 + 0.11726 (month)

The Concept of a Common Slope

Probit male = -1.001139 + 0.11726 (month)

Probit female = -0.41904 + 0.11726 (month)

Probit Z Score

F

Female Slope = 0.117

M

0.00 :

Male Slope = 0.117

-0.42 :

-1.02 :

0 1 2 3 4 5 6 7 8 9 10

Months Of Supervision

Notice that the difference between the two groups is in the intercepts: females = – 0.42, males = – 1.0. Males have a lower potential for success than females, regardless of the amount of supervision.